Here's a few thoughts on the question:
First, although this is probably obvious to everyone who's posted already, the region F must be bounded (if it is not $\mathbb{R}^2$). If not, then since it's convex, it must contain an infinite ray, and F must be contained in a half-space since it is convex. Take the projection onto the ray, then map the ray by arclength onto a spiral that can't live in any half-space, for example. This map then cannot be contained inside of F and is clearly length-decreasing.
So assume F is bounded. Then I think we may assume F is compact, by taking its closure. Any 1-Lipschitz map from F to $\mathbb{R}^2$ will extend to a 1-Lipschitz map from of the closure, and if the image lies in F, then its closure will lie in the closure. This probably doesn't help at all.
Edit: as Anton Petrunin has pointed out, the following argument is bogus: Now, the space of 1-Lipschitz maps to $\mathbb{R}^2$ is convex (and I think it is complete in the sup topology). Also, it's an easy exercise to see that any convex combination of 1-Lipschitz maps lying in isometric copies of F also lie in isometric copies of F (convex combinations of isometries give conformal affine maps with dilatation $\leq 1$). So to prove the claim for a given region F, we need "only" prove that extremal maps, i.e. ones which are not convex combinations of other maps, are contained in an isometric copy of F. I'm not sure if this helps, but there might be some literature on the convex structure of 1-Lipschitz maps which one could possibly exploit.
